Constructions and LociEdexcel GCSE Maths: Revision notes
Section 1
How do you construct triangles using ruler and compass?
To construct a triangle, you need three pieces of information (SSS, SAS, or ASA). The process uses only a ruler and compass — no protractor or measurements other than lengths.
Constructing a triangle given three sides (SSS):
- Draw the base line using a ruler
- Open the compass to the length of the first side
- From one end of the base, draw an arc
- Adjust compass to the second side length
- From the other end of the base, draw another arc
- Where the arcs meet is the third vertex
- Join all three points
Constructing a triangle given two sides and the included angle (SAS):
- Draw the first side using a ruler
- Measure the angle at one end (using compass arc method, not a protractor)
- Draw the second side at this angle
- Join the endpoints to complete the triangle
Key principle: All construction lines (arcs) must be left visible in your answer — examiners need to see your method.
Always leave your construction arcs visible — do not rub them out. The examiner marks the method, not just the final triangle. Mark key points clearly with a small cross or dot.
Students often erase their arcs after completing the construction. This loses marks because the examiner cannot verify your method. Keep all arc marks visible.
Section 2
How do you construct perpendicular bisectors and angle bisectors?
A perpendicular bisector passes through the midpoint of a line at 90°. An angle bisector divides an angle exactly in half.
To construct a perpendicular bisector of a line segment:
- Open the compass to more than half the line's length
- From one endpoint, draw an arc above and below the line
- Keep the same compass width and draw arcs from the other endpoint
- The two arcs intersect above and below the line
- Use a ruler to draw a line through these two intersection points
- This line is the perpendicular bisector — it crosses the original line at 90°
To construct an angle bisector:
- Place compass point at the angle's vertex
- Draw an arc that crosses both arms of the angle
- From each arc intersection point (on the angle's arms), draw two more arcs of equal radius
- These arcs meet at one point
- Draw a line from the vertex through this intersection point
- This line bisects (divides in half) the original angle
Remember: Both constructions use the compass to create equal distances, which is the geometric principle behind halving.
A perpendicular bisector is like finding the centre of a see-saw — equal distance from both ends and perfectly balanced. An angle bisector is like splitting a pizza slice exactly in half.
To bisect a 60° angle: place compass at the vertex, draw an arc crossing both arms, then from each arc intersection point draw equal arcs that meet at a point. A line through the vertex and this point creates two 30° angles.
Section 3
How do you construct perpendiculars from a point?
You may need to construct a perpendicular from a point to a line (the shortest distance) or a perpendicular from a point on a line.
Perpendicular from a point above/below a line:
- Place the compass point on the line and draw an arc that intersects the line at two points either side of where the perpendicular should touch
- Ensure both intersection points are equidistant from the original point
- Open the compass to more than half the distance between these two points
- From each intersection point, draw arcs on the opposite side of the line
- These arcs meet at a point
- Draw a line from the original point through this intersection point
- This line is perpendicular to the original line
Perpendicular at a point on a line:
- Place compass point at the point on the line
- Draw equal arcs on both sides along the line
- From each arc endpoint, draw arcs above (or below) the line with equal radius
- Draw a line through the intersection point and your original point
- This creates a 90° angle
This method is effectively a special case of the perpendicular bisector construction.
The perpendicular from a point to a line represents the shortest distance from that point to the line. Mark this clearly on your diagram — it is often required in loci problems.
Section 4
How do you construct standard angles (60° and 90°) using compass and ruler?
Two angles can be constructed using only compass and ruler without a protractor.
Constructing a 60° angle:
- Draw a line (this will be one arm of the angle)
- Place compass point at the point where the angle starts (the vertex)
- Draw an arc above the line — this arc can be any radius
- Without changing the compass width, place the compass point where the arc meets the line
- Draw another arc that intersects the first arc
- Draw a line from the vertex through the intersection of the two arcs
- The angle created is exactly 60°
Why 60°? An equilateral triangle has three 60° angles. When you construct arcs with equal radii, you create equal sides of an equilateral triangle.
Constructing a 90° angle:
- Construct the perpendicular bisector of a line segment, or use the perpendicular from a point method
- A perpendicular is always 90°
- Alternatively, construct a 60° angle, then bisect the remaining space to one side
Key principle: These constructions are based on geometric properties, not measurement.
To construct 60°: draw a base line, mark a point (vertex), draw an arc from the vertex with any radius, place compass on the arc's intersection with the line and draw another arc of the same radius to meet the first. Join the vertex to this intersection — the angle is 60°.
Students often change the compass width when constructing 60°. The radius must remain the same for both arcs — this is what creates the equilateral triangle principle.
Section 5
What are loci and how do you use them to solve problems?
A locus (plural: loci) is the set of all points satisfying a given condition. On an Edexcel GCSE paper, you must draw loci and find regions satisfying multiple conditions simultaneously.
Common loci:
| Condition | Locus | Shape |
|---|---|---|
| Equidistant from two points | Perpendicular bisector | Straight line |
| Equidistant from two lines | Angle bisector | Straight line |
| Fixed distance from a point | Circle | Curved |
| Fixed distance from a line | Parallel line (each side) | Two parallel lines |
| Equidistant from two intersecting lines | Both angle bisectors | Two perpendicular lines |
Solving loci problems with multiple conditions:
- Draw each locus separately (e.g., perpendicular bisector of AB)
- Draw all required loci on the same diagram
- Shade or identify the region satisfying all conditions (the intersection)
- Common problems: "Find points closer to A than B AND within 5 cm of line XY"
Key distinction: A locus is a line or curve; a region is an area bounded by loci.
Real-world applications: Finding locations for phone masts (equidistant from towns), safe zones around hazards (distance condition), or optimal meeting points.
Always show your construction lines and loci clearly. When asked to find a region, shade it carefully and make sure it satisfies all stated conditions — examiners check both the boundary lines and the shaded area.
Question: 'Find the locus of points equidistant from points A and B, and also within 3 cm of point A.' Answer: Draw the perpendicular bisector of AB (equidistant from both), then draw a circle of radius 3 cm centred at A. The locus region is where the perpendicular bisector intersects the circle.
Must Know
- Construction lines (arcs) must always be left visible — examiners mark your method, not just the final result
- Perpendicular bisector of a segment: draw equal-radius arcs from each endpoint; the line through the two arc intersections is perpendicular and passes through the midpoint
- Angle bisector: from the vertex, draw an arc crossing both arms; draw equal arcs from each crossing point; the line from the vertex through their intersection bisects the angle
- 60° angle is constructed using equal compass radii (based on equilateral triangle); 90° angle uses perpendicular methods (perpendicular bisector or perpendicular from a point)
- Locus is a line or curve of all points satisfying one condition; solve multi-condition loci problems by drawing all loci on the same diagram and identifying/shading the region satisfying every condition simultaneously
- Common loci: perpendicular bisector (equidistant from two points), angle bisector (equidistant from two lines), circle (fixed distance from a point), parallel lines (fixed distance from a line)
That's the notes covered.
Carry on to the next subtopic.